Question

[Dũng Senpai]

Cho \(\frac{1}{3}\le a,b,c\le3\)

Chứng minh:

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{7}{5}\)

Cảm ơn mọi người nha,mình cần gấp à.Bật mí là dùng phương pháp dồn biến ạ(ra biên hoặc toàn miền gì đó ạ)

Answer 1

[Nguyễn Linh Chi]

Đặt: f(a;b;c) =\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Vai trò của a, b, c là như nhau có thể giả sử: \(a=max\left\{a,b,c\right\}\)

Ta có: \(f\left(a;b;\sqrt{ab}\right)=\frac{a}{a+b}+\frac{b}{b+\sqrt{ab}}+\frac{\sqrt{ab}}{\sqrt{ab}+a}\)

\(=\frac{a}{a+b}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

Ta chứng minh:

\(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

+) Chứng minh: \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)

Xét : \(f\left(a;b;c\right)-f\left(a;b;\sqrt{ab}\right)=\frac{b}{b+c}+\frac{c}{a+c}-\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{b\left(a+c\right)\left(\sqrt{a}+\sqrt{b}\right)+c\left(b+c\right)\left(\sqrt{a}+\sqrt{b}\right)-2\sqrt{b}\left(b+c\right)\left(a+c\right)}{\left(b+c\right)\left(a+c\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{ab\sqrt{a}-ab\sqrt{b}+2bc\sqrt{a}-2ac\sqrt{b}+c^2\sqrt{a}-c^2\sqrt{b}}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\ge0\)vì a=max{a,b,c} => \(a\ge b\)

=> \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)(1)

+) Chứng minh:\(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

Xét: \(f\left(a;b;\sqrt{ab}\right)-\frac{7}{5}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{7}{5}\)\(=\frac{\frac{a}{b}}{\frac{a}{b}+1}+\frac{2}{\sqrt{\frac{a}{b}}+1}-\frac{7}{5}\)(2)

Đặt \(\sqrt{\frac{a}{b}}=x\left(đk:x\le3\right)\)Ta có: 

(2)=\(\frac{x^2}{x^2+1}+\frac{2}{x+1}-\frac{7}{5}\)\(=\frac{5x^3+5x^2+10x^2+10-7x^3-7x^2-7x-7}{5\left(x^2+1\right)\left(x+1\right)}\)

\(=\frac{-2x^3+8x^2-7x+3}{5\left(x^2+1\right)\left(x+1\right)}=\frac{\left(3-x\right)\left(2x^2-2x+1\right)}{5\left(x^2+1\right)\left(x+1\right)}\ge0\)

=> \(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)(3)

Từ (1); (3) => \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

"=" xảy ra <=> a=3; b=1/3; c=1 và các hoán vị